Past exam of the chemistry B course of the University of Cambridge / 2024 / Paper 2 / B7 / a

In the ionic assignment, both tetrahedral anions would require Mn(VII), $d^0$. Such a high oxidation state is strongly oxidizing and is best supported by strong bonding to oxygen. Short Mn–O bonds combine good orbital overlap with oxygen's resistance to further oxidation. This stabilizes permanganate relative to a hypothetical all-sulfide analogue.

sulfide is larger, more polarizable and much easier to oxidize than oxide. An attempted Mn(VII) sulfide would therefore combine an exceptionally oxidizing metal center with reducing ligands: internal redox can reduce manganese while oxidizing sulfur, for example toward S–S-bonded or elemental sulfur products. The larger sulfur donors also provide less favorable short multiple bonding to the small, highly charged manganese center. The oxygen compound supports Mn(VII); the sulfide combination is redox-incompatible. This predicts instability without requiring a specific unprovided decomposition product.

For an octahedral complex, strong axial sigma donation raises the metal–ligand antibonding $e_g$-derived orbitals. A hydride is a very strong sigma donor: its occupied $1s$ orbital overlaps effectively along a metal–ligand axis and has no filled perpendicular p orbitals to act as a conventional pi donor. It can therefore produce a large octahedral splitting by raising $e_g$ relative to $t_{2g}$.

Triphenylphosphine also gives sigma donation, but its useful acceptor orbitals additionally receive metal pi backbonding. This lowers the metal $t_{2g}$-derived levels and enlarges their separation from $e_g$. The acceptors have substantial P–C antibonding character; invoking a dominant phosphorus $3d$ acceptor is unnecessary. Thus hydride is strong-field principally through sigma bonding, while PPh${}_3$ combines sigma donation with pi acceptance. The spectrochemical series describes the resulting splitting, not just a ranking of ligand basicity.

The metal is Cr(II), $d^4$. A chloride ligand produces a relatively weak ligand field, so the octahedral center is a high-spin complex with $t_{2g}^3{e}_g^1$, $S=2$. The singly occupied $e_g$ pair is orbitally degenerate: $d_{z^2}$ and $d_{x^2-y^2}$ cannot retain identical occupancies once the distortion selects one. This is a strong setting for the Jahn-Teller effect.

On tetragonal elongation, the two axial Cr–Cl distances increase. Repulsion is reduced most for $d_{z^2}$, so it falls relative to $d_{x^2-y^2}$; $d_{xz},d_{yz}$ also fall relative to $d_{xy}$. The single $e_g$ electron occupies the lower $d_{z^2}$-derived level. The electronic stabilization offsets an elastic cost until a distorted equilibrium geometry is reached. The high-spin $d^4$ occupancy drives an elongated, rather than regular, octahedron. The three lower-set electrons remain one per orbital in this simple diagram; it is primarily the unequal $e_g$ occupancy that produces the strong distortion.

The all-carbonyl hydride, HCo(CO)${}_4$, is more acidic. Loss of ${\mathrm{H}}^{+}$ gives $[\mathrm{Co}(\mathrm{CO}{)}_4{]}^{-}$, whose negative charge is stabilized by pi backbonding into four carbonyl ligands. Replacing one CO by triphenyl phosphite reduces the net electron-withdrawing character of the ligand set relative to four carbonyls. The substituted hydride is more electron rich and its conjugate-base anion is less stabilized against protonation.Both conjugate bases have a conventional 18-electron count, so the difference is not a missing-electron argument. It is the relative stabilization of acid and base by the ligands in coordination chemistry. This concerns heterolytic deprotonation, not the homolytic Co–H bond-dissociation energy.